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On the Counteraction of "Lorentz" Torques

Abstract. The importance of considering mechanical forces, in addition to Lorentz forces, is illustrated by the initially puzzling results of the Trouton-Noble experiment.

Fig. 1 depicts charges q1 and q2 permanently at rest in inertial frame K. The electric field at q1, attributable to q2, has components , (1) . (2)

The Lorentz force on q1 thus has components , (3) . (4)

This force points from q1 to q2, and there is accordingly no torque around any point on the line between the charges.

Figure 1 Charges at Rest in K

Let inertial frame K’ move in the +x direction of K at speed v. Then the charges move in the –x’ direction of frame K’ at constant speed v. According to the field transformations the electric field components at q1, attributable to q2, are , (5) . (6)

Furthermore, from the perspective of K’, q2 engenders a magnetic field at q1: . (7)

The Lorentz force in K’ therefore has y’ component . (8)

The x’ component of the electric field in K’, on the other hand, equals the x component in K. Hence the Lorentz force x components are equal: . (9)

Since F’1x=F1x but F’1y<F1y, the line of action of the Lorentz force in K’ does not lie along the line between the charges. In K’ the Lorentz force evidently engenders a CCW torque around (say) the midpoint of the line between the charges. The only exception occurs when the line between the charges lies parallel to the x’ axis.

Of course there cannot be a zero torque in K and a nonzero torque in K’. For the charges move with a constant velocity in K’, and hence the system does not rotate in K’, quite as it does not in K. The solution to this conundrum lies in the fact that we have not considered the mechanical forces in K and K’. In K the mechanical force on q1 has components , (10) . (11)

In words, the charges are permanently at rest in K and hence (according to Newton) the mechanical forces must counteract the electric forces. The net force on each charge must be zero.

In K’ these forces transform to , (12) . (13)

Here again the mechanical forces counteract the Lorentz forces. And again, in K’ as in K, the Lorentz and mechanical forces on q1 sum to zero. Similarly for the total force on q2.

Historically it was believed that the charges would not rotate when at rest in the Maxwellian aether rest frame but, for all but the special orientation mentioned above, they would rotate when moving through the aether. Trouton and Noble attempted to detect such a rotation (owing to a presumed motion through the aether) by using oppositely charged capacitor plates. Evidently they considered neither the "mechanical" forces holding the plates apart, nor the "mechanical" forces holding the charges to the plates. They of course detected no rotation, regardless of the orientation of the charges. As in the Michelson-Morley experiment, there was at first considerable confusion over the failure to obtain the expected rotation.

Perhaps the lesson to be learned from Trouton-Noble is: "Always consider all of the forces ... particularly Lorentz and mechanical ... in any given problem." It is a lesson routinely neglected in electromagnetism texts. Failure to consider the mechanical force, however, may occasionally lead to spurious results a la Trouton-Noble, particularly when things are considered from more than one frame of reference!